1) Given the reaction below, calculate S univ at 298 K AND determine whether the
ID: 509203 • Letter: 1
Question
1) Given the reaction below, calculate Suniv at 298 K AND determine whether the reaction is spontaneous.
4 Fe (s) + 3 O2 (g) 2 Fe2O3 (s)
2) Given the reaction below, calculate Grxn at 298 K AND determine whether the reaction is spontaneous.
N2 (g) + O2 (g) 2 NO (g)
3)Which statement is true?
Endothermic reactions may be spontaneous at low temperature if there is a positive change in entropy.
Exothermic reactions are always spontaneous.
Exothermic reactions are always nonspontaneous if there is a positive change in entropy.
Endothermic reactions may be spontaneous at high temperatures if there is a positive change in entropy.
Endothermic reactions are always spontaneous if there is a negative change in entropy.
4) Given the reaction below, at what temperature range will the reaction be spontaneous?
N2O4 (g) 2 NO2 (g)
5) Given reaction and initial concentrations below, what is G and which direction of the reaction is spontaneous.
H2 (g) + I2 (g) 2 HI (g)
[H2]i = 0.10 M, [I2]i = 0.20 M, [HI]i = 0.30 M
6) Given the reaction below, calculate K at 298 K.
I2 (s) I2 (g)
Endothermic reactions may be spontaneous at low temperature if there is a positive change in entropy.
Exothermic reactions are always spontaneous.
Explanation / Answer
1) Use the following table to answer the question:
Substance (state)
fH0 (kJ/mol)
S0(J/mol.K)
Fe (s)
0
27.3
O2 (g)
0
205.0
Fe2O3 (s)
-824.2
87.4
Calculate the entropy change for the reaction:
rS0 = [(2 mole)*S0(Fe2O3, s)] – [(4 mole)*S0(Fe, s) + (3 mole)*S0(O2, g)] = [(2 mole)*(87.4 J/mol.K)] – [(4 mole)*(27.3 J/mol.K) + (3 mole)*(205.0 J/mol.K)] = (174.8 J/K) – (109.2 + 615) J/K = (174.8 – 724.2) J/K = -549.4 J/K.
To calculate the entropy change of the universe, calculate the entropy change of the surroundings. To obtain rS0 (surr), obtain rH0 (sys).
rH0 = [(2 mole)*fH0(Fe2O3, s)] – [(4 mole)*fH0(Fe, s) + (3 mole)*fH0(O2, g)] = [(2 mole)*(-824.2 kJ/mol)] – [(4 mole)*0 + (3 mole)*0] = -1648.4 kJ.
rS0 (surr) = -rH0/T = -(-1648.4 kJ)/(298 K) = 5.5315 kJ/K = (5.5315 kJ/K)*(1000 J/1 kJ) = 5531.5 J/K.
S (univ) = rS0 (sys) + rS0 (surr) = (-549.4 J/K) + (5531.5 J/K) = 4982.1 J/K (ans)
Since S (univ) > 0, the reaction is spontaneous at 298 K.
2) Use the following table to answer the question:
Substance (state)
fH0 (kJ/mol)
S0(J/mol.K)
N2 (g)
0
191.5
O2 (g)
0
205.0
NO (g)
90.2
210.7
Calculate the entropy change for the reaction:
rS0 = [(2 mole)*S0(NO, g)] – [(1 mole)*S0(N2, g) + (1 mole)*S0(O2, g)] = [(2 mole)*(210.7 J/mol.K)] – [(1 mole)*(191.5 J/mol.K) + (1 mole)*(205.0 J/mol.K)] = (421.4 J/K) – (191.5 + 205.0) J/K = (421.4 – 396.5) J/K = 24.9 J/K.
Calculate the enthalpy change for the reaction, rH0.
rH0 = [(2 mole)*fH0(NO, g)] – [(1 mole)*fH0(N2, g) + (1 mole)*fH0(O2, g)] = [(2 mole)*(90.2 kJ/mol)] – [(1 mole)*0 + (1 mole)*0] = 90.2 kJ.
Calculate the free energy change for the reaction at 298 K
rG0 = rH0 – T*rS0 = (90.2 kJ) – (298 K)*(24.9 J/K) = (90.2 kJ) – (7420.2 J) = (90.2 kJ) – (7420.2 J)*(1 kJ/1000 J) = (90.2 kJ) – (7.4202 kJ) = 82.7798 kJ 82.8 kJ
Since rG0 > 0, the reaction is non-spontaneous at 298 K (ans).
Substance (state)
fH0 (kJ/mol)
S0(J/mol.K)
Fe (s)
0
27.3
O2 (g)
0
205.0
Fe2O3 (s)
-824.2
87.4
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